Lecture 11

• One-way analysis of variance (Chapter 15.2)

Review: Relat. between One-Sided Hypothesis Tests and CIs

• Suppose we are given a CI for

• For the one-sided hypothesis test versus at significance level , we can conclude– We reject the null hypothesis if and does

not belong to the confidence interval– We do not reject the null hypothesis if either

or belongs to the confidence interval.

%100)1(

00 : H

01 : H 2/

0x0

0x

0

Review: CIs for Monotonic Functions of Parameters

• A function f(x) is monotonic if it moves in one direction as its argument increases.

• Suppose that we have a CI for a parameter and that we want to find a CI for the parameter .

• If f is monotonically increasing, the CI is

. If f is monotonically decreasing, the CI is

)( , UL

)(f

))(),(( UL ff ))(),(( LU ff

Review of one-way ANOVA

• Objective: Compare the means of K populations of interval data based on independent random samples from each.

• H0: • H1: At least two means differ• Notation: xij – ith observation of jth sample;

- mean of the jth sample; nj – number of observations in jth sample; - grand mean of all observations

K 21

jx

x

• The marketing manager for an apple juice manufacturer needs to

decide how to market a new product. Three strategies are considered,

which emphasize the convenience, quality and low price of product

respectively.

• An experiment was conducted as follows:

• In three cities an advertisement campaign was launched .

• In each city only one of the three characteristics (convenience,

quality, and price) was emphasized.

• The weekly sales were recorded for twenty weeks following

the beginning of the campaigns.

Example 15.1

Rationale Behind Test Statistic

• Two types of variability are employed when testing for the equality of population means– Variability of the sample means– Variability within samples

• Test statistic is essentially (Variability of the sample means)/(Variability within samples)

The rationale behind the test statistic – I

• If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean).

• If the alternative hypothesis is true, at least some of the sample means would differ.

• Thus, we measure variability between sample means.

• The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean.

This sum is called the

Sum of Squares for Treatments

SSTIn our example treatments arerepresented by the differentadvertising strategies.

Variability between sample means

2k

1jjj )xx(nSST

There are k treatments

The size of sample j The mean of sample j

Sum of squares for treatments (SST)

Note: When the sample means are close toone another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H1.

• Solution – continuedCalculate SST

2k

1jjj

321

)xx(nSST

65.608x00.653x577.55x

= 20(577.55 - 613.07)2 + + 20(653.00 - 613.07)2 + + 20(608.65 - 613.07)2 == 57,512.23

The grand mean is calculated by

k21

kk2211

n...nnxn...xnxn

X

Sum of squares for treatments (SST)

Is SST = 57,512.23 large enough to reject H0 in favor of H1?

Large compared to what?

Sum of squares for treatments (SST)

20

25

30

1

7

Treatment 1 Treatment 2 Treatment 3

10

12

19

9

Treatment 1Treatment 2Treatment 3

20

161514

1110

9

10x1

15x2

20x3

10x1

15x2

20x3

The sample means are the same as before,but the larger within-sample variability makes it harder to draw a conclusionabout the population means.

A small variability withinthe samples makes it easierto draw a conclusion about the population means.

• Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means.

• Therefore, even though sample means may markedly differ from one another, SST must be judged relative to the “within samples variability”.

The rationale behind test statistic – II

• The variability within samples is measured by adding all the squared distances between observations and their sample means.

This sum is called the

Sum of Squares for Error

SSEIn our example this is the sum of all squared differencesbetween sales in city j and thesample mean of city j (over all the three cities).

Within samples variability

• Solution – continuedCalculate SSE

Sum of squares for errors (SSE)

k

jjij

n

i

xxSSE

sss

j

1

2

1

23

22

21

)(

24.670,811,238,700.775,10

(n1 - 1)s12 + (n2 -1)s2

2 + (n3 -1)s32

= (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50

Is SST = 57,512.23 large enough relative to SSE = 506,983.50 to reject the null hypothesis that specifies that all the means are equal?

Sum of squares for errors (SSE)

To perform the test we need to calculate the mean squaresmean squares as follows:

The mean sum of squares

Calculation of MST - Mean Square for Treatments

12.756,2813

23.512,571

k

SSTMST

Calculation of MSEMean Square for Error

45.894,8360

50.983,509

kn

SSEMSE

And finally the hypothesis test:

H0: 1 = 2 = …=k

H1: At least two means differ

Test statistic:

R.R: F>F,k-1,n-k

MSEMST

F

The F test rejection region

The F test

Ho: 1 = 2= 3

H1: At least two means differ

Test statistic F= MST MSE= 3.2315.3FFF:.R.R 360,13,05.0knk 1

Since 3.23 > 3.15, there is sufficient evidence to reject Ho in favor of H1, and argue that at least one of the mean sales is different than the others.

23.317.894,812.756,28

MSEMST

F

Required Conditions for Test

• Independent simple random samples from each population

• The populations are normally distributed (look for extreme skewness and outliers, probably okay regardless if each ).

• The variances of all the populations are equal (Rule of thumb: Check if largest sample standard deviation is less than twice the smallest standard deviation)

30jn

ANOVA Table – Example 15.1Analysis of Variance

Source DF Sum of Squares

Mean Square

F Ratio Prob > F

City 2 57512.23 28756.1 3.2330 0.0468

Error 57 506983.50 8894.4    

C. Total

59 564495.73      

Model for ANOVA

• = ith observation of jth sample• • is the overall mean level, is the differential

effect of the jth treatment and is the random error in the ith observation under the jth treatment. The errors are assumed to be independent, normally distributed with mean zero and variance The are normalized:

ijX

ijjijX

j

ij

2

j

K

j j10

Model for ANOVA Cont.

• The expected response to the jth treatment is

• Thus, if all treatments have the same expected response (i.e., H0 : all populations have same mean), . In general, is the difference between the means of population j and j’.

• Sums of squares decomposition: SS(Total)=SST+SSE

jijXE )(

Kjj ,,1for ,0 'jj

Relationship between F-test and t-test for two samples

• For comparing two samples, the F-statistic equals the square of the t-statistic with equal variances.

• For two samples, the ANOVA F-test is equivalent to testing versus .

210 : H

211 : H

Practice Problems

• 15.16, 15.22, 15.26

• Next Time: Chapter 15.7 (we will return to Chapters 15.3-15.5 after completing Chapter 15.7).